JEE Main 2016 · 9 Apr · Shift 1 · Chemistry Q49

ChemistrySingle correct+4 / -1Question 49 of 90

Identify the correct trend given below :

(Atomic No. = Ti : 22, Cr : 24 and Mo : 42)

Δo of

\ce{[Cr(H2O)6]^2+}

\ce{[Mo(H2O)6]^2+}

and Δo of

\ce{[Ti(H2O)6]^3+}

\ce{[Ti(H2O)6]^2+}

Δo of

\ce{[Cr(H2O)6]^2+}

\ce{[Mo(H2O)6]^2+}

and Δo of

\ce{[Ti(H2O)6]^3+}

\ce{[Ti(H2O)6]^2+}

Δo of

\ce{[Cr(H2O)6]^2+}

\ce{[Mo(H2O)6]^2+}

and Δo of

\ce{[Ti(H2O)6]^3+}

\ce{[Ti(H2O)6]^2+}

Δo of

\ce{[Cr(H2O)6]^2+}

\ce{[Mo(H2O)6]^2+}

and Δo of

\ce{[Ti(H2O)6]^3+}

\ce{[Ti(H2O)6]^2+}

Answer and worked solution

Official answer: C

Δo increases down a group (4d > 3d), so $\ce{[Cr(H2O)6]^2+}$ < $\ce{[Mo(H2O)6]^2+}$ , and with the oxidation state, so $\ce{Ti^3+}$ > $\ce{Ti^2+}$ .

JEE Main 2016 · 9 Apr · Shift 1, Chemistry question 49